Chapter 1.4, Problem 11E

(a)

To determine

To find: The graph of the parametric equations, initial and terminal points. Also, indicate the direction in which the curve is traced.

Answer to Problem 11E

The graph of the parametric equations is shown in figure (1), the initial point is $\left(-1,0\right)$ and terminal point is $\left(0,1\right)$.

Explanation of Solution

Given information: The equations are $x=\cos \left(\pi -t\right)$ and $y=\sin \left(\pi -t\right)$ , $0\le t\le \pi$.

Calculation:

Use the following step to graph the parametric equations by graphing calculator.

Step 1: First press the “ON” button graphical calculator.

Step 2: Press the $\boxed{\text{MODE}}$ button, under FUNC and choose PAR option.

Step 3: Press $\boxed{\text{Y=}}$ , and set the parametric equations equal to ${\text{X}}_{\text{1T}}$ and ${\text{Y}}_{\text{1T}}$.

Step 4: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-3,3\right]$ by $\left[-2,2\right]$ as shown below.

  

Figure (1)

If $t=0$ , then $x=-1$ and $y=0$ . So the initial point is $\left(-1,0\right)$.

If $t=\pi$ , then $x=1$ and $y=0$ . So, the terminal point is $\left(1,0\right)$.

So, the initial point is $\left(-1,0\right)$ and terminal point is $\left(1,0\right)$.

Therefore, the graph of the parametric equations is shown in figure (1), the initial point is $\left(-1,0\right)$ and terminal point is $\left(1,0\right)$.

(b)

To determine

To find: The Cartesian equation for a curve that contains the parameterized curve and the portion of the graph of the Cartesian equation that is traced by the parameterized curve.

Answer to Problem 11E

The Cartesian equation for a curve that contains the parameterized curve is $x^2+y^2=1$ and parameterized curve traces the upper half of the circle defined by $x^2+y^2=1$.

Explanation of Solution

Given information: The equations are $x=\cos \left(\pi -t\right)$ and $y=\sin \left(\pi -t\right)$ , $0\le t\le \pi$.

Calculation:

Add the equations $x=\sin \left(2\pi t\right)$ , and $y=\cos \left(2\pi t\right)$ after squaring them.

  $\begin{array}{c}{x}^2+y^2={\sin }^2\left(\pi -t\right)+{\cos }^2\left(\pi -t\right)\\ =1\end{array}$

As shown in the graph, the parameterized curve traces the upper half of the circle defined by $x^2+y^2=1$.

Therefore, the Cartesian equation for a curve that contains the parameterized curve is $x^2+y^2=1$ and parameterized curve traces the upper half of the circle defined by $x^2+y^2=1$.